## College Physics (4th Edition)

(a) If a wheel is rolling by itself, a significant fraction of the total kinetic energy is rotational kinetic energy. Therefore, we can not assume that the wheel is sliding without friction. (b) The fraction of the total kinetic energy that is the rotational kinetic energy is $0.0174$
(a) If a wheel is rolling by itself, a significant fraction of the total kinetic energy is rotational kinetic energy. Therefore, we can not assume that the wheel is sliding without friction since this would not be accurate. (b) We can find an expression for the total rotational kinetic energy of the four wheels: $KE_{rot} = 4\times \frac{1}{2}I\omega^2$ $KE_{rot} = (2)(I)(\frac{v}{r})^2$ $KE_{rot} = (2)(0.705~kg~m^2)(\frac{v}{0.35~m})^2$ $KE_{rot} = (11.51~kg)~v^2$ We can find an expression for the translational kinetic energy: $KE_{tr} = \frac{1}{2}mv^2$ $KE_{tr} = \frac{1}{2}(1300~kg)~v^2$ $KE_{tr} = (650~kg)~v^2$ We can find the fraction of the total kinetic energy that is the rotational kinetic energy of the wheels: $\frac{KE_{rot}}{KE} = \frac{KE_{rot}}{KE_{tr}+KE_{rot}}$ $\frac{KE_{rot}}{KE} = \frac{(11.51~kg)~v^2}{(650~kg)~v^2+(11.51~kg)~v^2}$ $\frac{KE_{rot}}{KE} = 0.0174$ The fraction of the total kinetic energy that is the rotational kinetic energy of the wheels is $0.0174$.